Advanced book on Mathematics Olympiad

624 Geometry and Trigonometry

(c) Substituting the coordinates ofLin the equation of the circle yields

(ac+ 4 p^2 )(a−b)(c−b)= 0.

Sincea  =b  =c, we must haveac=− 4 p^2. Thus thex-coordinate ofNis−p, showing

that this point is on the directrix.

(d) The condition forFto be onACis 4p^2 +ac=0, in which caseNis on the

directrix. The slope ofBFism=b (^24) −pb 4 p 2. The orthogonality condition is
4 pb
b^2 − 4 p^2

·

4 p

c+a

=− 1 ,

which is equivalent to

(b^2 − 4 p^2 )(c+a)+ 16 p^2 b= 0.

The locus is obtained by eliminatinga, b, cfrom the equations

4 px−(c+a)y+ca= 0 ,

y=b,

4 p^2 +ac= 0 ,

(b^2 − 4 p^2 )(c+a)+ 16 p^2 b= 0.

The answer is the cubic curve

(y^2 − 4 p^2 )x+ 3 py^2 + 4 p^3 = 0.

(TheMathematics GazetteCompetition, Bucharest, 1938)

611.An equilateral triangle can be inscribed in any closed, non-self-intersecting curve,

therefore also in an ellipse. The argument runs as follows. Choose a pointAon the

ellipse. Rotate the ellipse aroundAby 60◦. The image of the ellipse through the rotation

intersects the original ellipse once inA, so it should intersect it at least one more time.

LetBan be intersection point different fromA. Note thatBis on both ellipses, and its

preimageCthrough rotation is on the original ellipse. The triangleABCis equilateral.

A square can also be inscribed in the ellipse. It suffices to vary an inscribed rectangle

with sides parallel to the axes of the ellipse and use the intermediate value property.

Let us show that these are the only possibilities. Up to a translation, a rotation, and

a dilation, the equation of the ellipse has the form

x^2 +ay^2 =b, witha, b > 0 ,a   = 1.

Assume that a regularn-gon,n≥5, can be inscribed in the ellipse. Its vertices(xi,yi)
satisfy the equation of the circumcircle: